Twin Paradox Problem: Do Twins Age Differently?

[Rishavutkarsh]

twin paradox problem!?

ok it's about the twin paradox the guy on spaceship goes at 90% of C (say) but since both the twins see each other's clocks equally slow and if one sees other at 90%C then the other one also sees his twin at 90%C in opposite direction. then how on returning the twin on spaceship is younger?
is there something I am missing?

 

[HallsofIvy]

That's the whole point of the "twin paradox" and why it is called a "paradox"- the symmetry.

But the "returning" part means there cannot be symmetry. In order for the twin who goes out an comes back to be able to return, he must accelerate. While velocity is relative, acceleration is not.

 

[harrylin]

ok it's about the twin paradox the guy on spaceship goes at 90% of C (say) but since both the twins see each other's clocks equally slow and if one sees other at 90%C then the other one also sees his twin at 90%C in opposite direction. then how on returning the twin on spaceship is younger?
is there something I am missing?

Did you read some of the earlier threads on the same topic? See a few links at the bottom of this page. Especially this one: https://www.physicsforums.com/showthread.php?t=374408
Any question remains?

 

[TrickyDicky]

I read a few twin paradox threads, including the one suggested in this thread, and I am not sure a consensus has been reached about it's solution. It seems half the people claims the key is in the acceleration of the traveling twin, and the other half says acceleration has nothing to do, (not to mention those that want to solve it introducing GR).
So what is the official PF position?:-/

 

[ghwellsjr]

ok it's about the twin paradox the guy on spaceship goes at 90% of C (say) but since both the twins see each other's clocks equally slow and if one sees other at 90%C then the other one also sees his twin at 90%C in opposite direction. then how on returning the twin on spaceship is younger?
is there something I am missing?

You are only half correct. When the traveling twin departs at 90%c, they both do see each others clock equally slow--a factor of 0.2294 times their own. But this is true only for the outbound portion of the trip. Things are different for the inbound portion of the trip. As soon as the traveling twin turns around, he immediately sees the Earth twin's clock going fast--4.359 times his own. Since he spends an equal amount of time going out as coming in, you can easily calculate how much of a difference there will be in the amount the two twins aged by simply taking an average of the two factors. The average of 0.2294 and 4.359 is 2.2942 so however much the traveling twin aged during the trip, his Earth twin will age 2.2942 times as much. Simple, isn't it?

 

[Rishavutkarsh]

You are only half correct. When the traveling twin departs at 90%c, they both do see each others clock equally slow--a factor of 0.2294 times their own. But this is true only for the outbound portion of the trip. Things are different for the inbound portion of the trip. As soon as the traveling twin turns around, he immediately sees the Earth twin's clock going fast--4.359 times his own. Since he spends an equal amount of time going out as coming in, you can easily calculate how much of a difference there will be in the amount the two twins aged by simply taking an average of the two factors. The average of 0.2294 and 4.359 is 2.2942 so however much the traveling twin aged during the trip, his Earth twin will age 2.2942 times as much. Simple, isn't it?

well it surely is but isn't it said that a person sees the clock fastest which is at rest compared to him so can he really see the other clock running 4.359 (or whatever)
also is there droppler's effect also involved in this in any way?

 

[Rishavutkarsh]

That's the whole point of the "twin paradox" and why it is called a "paradox"- the symmetry.

But the "returning" part means there cannot be symmetry. In order for the twin who goes out an comes back to be able to return, he must accelerate. While velocity is relative, acceleration is not.

that's what is the point of dispute some say it's due to acceleration while others say that it has nothing to do

 

[harrylin]

I read a few twin paradox threads, including the one suggested in this thread, and I am not sure a consensus has been reached about it's solution. It seems half the people claims the key is in the acceleration of the traveling twin, and the other half says acceleration has nothing to do, (not to mention those that want to solve it introducing GR).
So what is the official PF position?:-/

The official PF position?

In the multitude of answers some subtle points may have gone unnoticed, and while there was no disagreement about this issue as far as I am aware, I can understand the need for clarification.

1. As remarked in the first full discussion of the "twin" scenario in the literature (from both points of view, and in which the stay-at-home is approximately at constant inertial motion), the symmetry is broken by a change in velocity of the traveler.

2. The acceleration that is required for this change in velocity has itself no effect on clock rate; clock rate is calculated as function of the clock's speed.

If not well understood, from point 1 one may think that acceleration itself causes a difference in clock rate, but that is wrong; or from point 2 one may think that the traveler may be considered the whole time at rest in an inertial frame, but that is wrong too.

Depending on which reference frame you choose, speeds appear differently; and if one chooses two different reference frames for the traveler (typically the ones in which the traveler is in rest during different periods), then one has to do a Lorentz transformation when one makes the switch. And different frames attribute different times to distant events.

 

[Jeronimus]

... While velocity is relative, acceleration is not...

I am not sure i agree fully with this.

Taking the situation, Earth with the twins at the same place with synced clocks, i can see there being an observer which is also at the same place shortly before the acceleration AND who is within an inertial reference frame at rest which after the acceleration will see both rockets moving away of each other at the same velocity seen from his point of view. He would see both twins accelerating symmetrically.

If such a thing is possible, then how is acceleration not relative? It is not symmetric for all inertial reference systems. Does breaking the symmetry in some inertial reference frames imply that something is absolute rather than relative? As i understand it, it does not.

 

[ghwellsjr]

I am not sure i agree fully with this.

Taking the situation, Earth with the twins at the same place with synced clocks, i can see there being an observer which is also at the same place shortly before the acceleration AND who is within an inertial reference frame at rest which after the acceleration will see both rockets moving away of each other at the same velocity seen from his point of view. He would see both twins accelerating symmetrically.

If such a thing is possible, then how is acceleration not relative? It is not symmetric for all inertial reference systems. Does breaking the symmetry in some inertial reference frames imply that something is absolute rather than relative? As i understand it, it does not.

No, you are wrong. The third inertial observer that you describe will see both twins prior to acceleration traveling in the same direction at the same speed with respect to him. Then he will see just one twin accelerate so that he ends up traveling in the opposite direction at the same speed as before. The other twin remains inertial.

So before acceleration, he sees both twins traveling towards him at a constant speed. Then when they get to him, one of the suddenly reverses direction while the other one continues on just as before and then he sees both of the traveling away from him at the same speed.

But what has this to do with the topic of this thread?

 

[ghwellsjr]

that's what is the point of dispute some say it's due to acceleration while others say that it has nothing to do

Your problem is that you misrepresented what the twins will see. They will not both see the other ones clock going slower than their own as I pointed out in post #5. You need to understand that every one of the other explanations will agree on what the twins see of the other ones clock. There is no dispute about what the explanations do, all the explanations work and agree on what the twins see which is what you asked about. The only dispute is about how some people insist that their personally favorite explanation is better than all the others but anyone who understands relativity will agree that all these explanations are equally valid.

You need to quit misrepresenting the Twin Paradox and the different explanations of the Twin Paradox.

 

[Jeronimus]

removed for edit

 

[TrickyDicky]

The official PF position?

In the multitude of answers some subtle points may have gone unnoticed, and while there was no disagreement about this issue as far as I am aware, I can understand the need for clarification.

1. As remarked in the first full discussion of the "twin" scenario in the literature (from both points of view, and in which the stay-at-home is approximately at constant inertial motion), the symmetry is broken by a change in velocity of the traveler.

2. The acceleration that is required for this change in velocity has itself no effect on clock rate; clock rate is calculated as function of the clock's speed.

If not well understood, from point 1 one may think that acceleration itself causes a difference in clock rate, but that is wrong; or from point 2 one may think that the traveler may be considered the whole time at rest in an inertial frame, but that is wrong too.

Depending on which reference frame you choose, speeds appear differently; and if one chooses two different reference frames for the traveler (typically the ones in which the traveler is in rest during different periods), then one has to do a Lorentz transformation when one makes the switch. And different frames attribute different times to distant events.

Ok, this is helpful. My own understanding is along these lines.
To me the asymmetry comes simply from the fact that in SR inertial frames are preferred over non-inertial ones, and the inertial one is the one that always experiments the longest proper time.
What is intriguing IMO is that by introducing the two different kind of frames and thus the asymmetry that dissolves the
paradox, and in the scheme that considers accelerations for the traveling twin as instantaneous, it introduces a way to see velocities as absolute, something that is mathematically impossible when using only inertial frames in SR.

 

[Jeronimus]

No, you are wrong. The third inertial observer that you describe will see both twins prior to acceleration traveling in the same direction at the same speed with respect to him. Then he will see just one twin accelerate so that he ends up traveling in the opposite direction at the same speed as before. The other twin remains inertial.

So before acceleration, he sees both twins traveling towards him at a constant speed. Then when they get to him, one of the suddenly reverses direction while the other one continues on just as before and then he sees both of the traveling away from him at the same speed.

But what has this to do with the topic of this thread?

I was imagining an observer which is at free fall i guess, for whom both rockets would look like they are accelerating. Not sure if this can be considered an inertial frame of reference. Anyway, i will not further this as it gets too complicated for me.

 

[harrylin]

Ok, this is helpful. My own understanding is along these lines.
To me the asymmetry comes simply from the fact that in SR inertial frames are preferred over non-inertial ones, and the inertial one is the one that always experiments the longest proper time.

More or less so: as determined in an inertial frame, a clock that is at rest accumulates more proper time than one that is moving, and that was already found in Einstein's 1905 paper. And as a matter of fact, the first "twin" illustration with observations from both sides was part of an article that explained these things:
- https://en.wikisource.org/wiki/The_Evolution_of_Space_and_Time

What is intriguing IMO is that by introducing the two different kind of frames and thus the asymmetry that dissolves the paradox, and in the scheme that considers accelerations for the traveling twin as instantaneous, it introduces a way to see velocities as absolute, something that is mathematically impossible when using only inertial frames in SR.

Actually, as you can see in the above link, it was the very purpose of that example to illustrate that acceleration is "absolute" in an observational sense (not clear what you mean with "a way to see velocities as absolute", except if you basically mean the same as him). And it became known as a "paradox" because Einstein tried to make accelerations also "relative" in the development of GR.

 

[PeterDonis]

To me the asymmetry comes simply from the fact that in SR inertial frames are preferred over non-inertial ones, and the inertial one is the one that always experiments the longest proper time.

First, a small quibble over terminology: it's not the "inertial frame" that experiences the longest proper time, it's the observer in *inertial motion*. (And strictly speaking, "longest proper time" means "longest proper time between two specific events.)

The more substantive issue I have with your way of putting it is that it does not generalize to curved spacetimes. Consider, for example, the following two observers: one on the International Space Station, and another "hovering" over the Earth at the same altitude as the ISS, but not orbiting the Earth (i.e., "hovering" motionless with respect to the distant stars). The first observer is inertial, the second is not; but if we pick two successive events where the two observers meet (between which the ISS completes one orbit), the second observer (the non-inertial one) will experience more elapsed proper time between them than the first (the inertial one).

So IMO a more general "solution" to the "paradox" is needed, and that is to simply understand that "elapsed time" is length along a curve, and that different curves between the same pair of points can have different lengths. The two twins in the original "paradox" travel on different curves between the same pair of points, so they experience different "lengths" (elapsed proper times). Once you're over that hurdle, figuring out which of the lengths is longer is just calculation; but the real hurdle is getting people to understand that "elapsed proper time" is just length along a curve. That concept generalizes easily to *any* timelike curve.

 

[TrickyDicky]

Actually, as you can see in the above link, it was the very purpose of that example to illustrate that acceleration is "absolute" in an observational sense (not clear what you mean with "a way to see velocities as absolute", except if you basically mean the same as him).

I was thinking of your point 2. above. It introduces a way to assign velocities and rest preferentially to the twins based only on speed.

 

[Nugatory]

I read a few twin paradox threads, including the one suggested in this thread, and I am not sure a consensus has been reached about its solution. It seems half the people claims the key is in the acceleration of the traveling twin, and the other half says acceleration has nothing to do, (not to mention those that want to solve it introducing GR).
So what is the official PF position?:-/

I think the lack of consensus is not so much about the resolution of the paradox as it is about the best way to explain it to someone who doesn't understand relativity at the level of the Minkowski geometry. (If you do understand the Minkowski geometry there isn't any paradox to explain, it all just makes sense).

 

[phyti]

If each twin experiences an acceleration, they may accumulate the same or different amounts of time at reunion. This eliminates acceleration as a causative factor.

 

[TrickyDicky]

I think the lack of consensus is not so much about the resolution of the paradox as it is about the best way to explain it to someone who doesn't understand relativity at the level of the Minkowski geometry.

Could be.

If you do understand the Minkowski geometry there isn't any paradox to explain, it all just makes sense.

I'm not sure it is so straight forward, the introduction of non-inertial frames (curvilinear coordinates) in a flat space can lead to confusion( as the long countless threads on the paradox testify) even to people who is aware of the properties of Minkowski space.

 

[TrickyDicky]

If each twin experiences an acceleration, they may accumulate the same or different amounts of time at reunion. This eliminates acceleration as a causative factor.

If both twins experience change of velocity (so both are noninertial), how do you decide which one is older/younger at reunion?

 

[zonde]

2. The acceleration that is required for this change in velocity has itself no effect on clock rate; clock rate is calculated as function of the clock's speed.

Acceleration have coordinate effect on clock reading. That's because acceleration changes the hyperplane of now.
Imagine two observers at the same spot moving at the same speed in respect to Earth but in opposite directions (toward/away). While they see exactly the same picture of Earth they will assign different readings to the "now" of Earth clock.

 

[Nugatory]

Acceleration have coordinate effect on clock reading. That's because acceleration changes the hyperplane of now.
Imagine two observers at the same spot moving at the same speed in respect to Earth but in opposite directions (toward/away). While they see exactly the same picture of Earth they will assign different readings to the "now" of Earth clock.

Ah... where's the acceleration in this example?

 

[TrickyDicky]

We should keep in mind we are in the domain of SR, that is, flat Minkowski spacetime, regardless of the use of different type of reference frames/coordinates the physics should not change.

 

[Jeronimus]

I read a few twin paradox threads, including the one suggested in this thread, and I am not sure a consensus has been reached about it's solution. It seems half the people claims the key is in the acceleration of the traveling twin, and the other half says acceleration has nothing to do, (not to mention those that want to solve it introducing GR).
So what is the official PF position?:-/

As i see it,

when the twins are at rest on earth, the acceleration which changes which frame the leaving twin is at rest in, is not the cause of him aging less.

One could imagine, that before the twin leaves earth, he places a long line of synced red colored clocks in front and behind him. Let's assume those clocks also have negative counters, with the acceleration taking place when the clock is at zero.

The frame Earth and the twins are at rest in before the acceleration takes place is frame A.
Frame B is the frame the leaving twin will be at rest in after the acceleration.

Now assume that in frame B there is a guy named Bob, who also placed synced blue colored clocks in front and behind him. Just when Bob passes by the twins at vrel, the leaving twin accelerates instantaneous. Bob has set the clocks in such a way, that just when he passes by the twins, the clocks are at zero seen from within his rest frame.

Bob will accelerate instantaneous just when he passes by the twins, when all the blue clocks in his frame show zero.
The staying twin and Bob are now at the same place at rest in frame A, while the leaving twin is at rest in frame B moving at vrel = 0.9c relative to Bob/staying twin.


The leaving twin now at rest in frame B, will see the red clocks display higher counts the further away they are in front of him, and lower counts the further away they are at the back of his rocket.


Bob will see the equivalent, concerning the blue clocks he placed in his initial frame B BEFORE accelerating. At rest in frame A AFTER accelerating, Bob will see blue clocks in front of him are now showing higher counts, while behind him they show lower count. Higher/lower the further away.


Assume Bob and the twins are of the same age when the acceleration (instantaneous) events take place at t=0.


Bob is basically in the same position as the staying twin now, but went through the equivalent acceleration process the leaving twin went through.

Neither Bob nor the leaving twin seem to be any special in this regard. The situation to me looks symmetrical.
Therefore it is not the LOCAL acceleration at t=0 which is the cause of the age difference.

The acceleration/accelerations which occur non-local are the cause for the age difference once they meet up again.

In fact, instead of the leaving twin returning, the staying twin or Bob could change his mind, and decide to accelerate towards the leaving twin. In that case, Bob/the staying twin would have aged less.

The initial acceleration of the leaving twin made no difference in the aging. It was necessary however, to get a distance between the twins, allowing for the combination of acceleration and distance to cause the difference in aging.

So no, acceleration is not the cause of the difference in aging, but the distance to each other combined with acceleration which makes this happen.

 

[TrickyDicky]

Jeronimus, I'm aware from the start that the acceleration people talks about in relation with the TP is the distant "turn around" acceleration.

 

[harrylin]

Acceleration have coordinate effect on clock reading. That's because acceleration changes the hyperplane of now.
Imagine two observers at the same spot moving at the same speed in respect to Earth but in opposite directions (toward/away). While they see exactly the same picture of Earth they will assign different readings to the "now" of Earth clock.

Only if and when they switch the inertial reference frames that they use, as I mentioned. For example the astronauts in the international space-station likely stick to the ECI frame instead of continuously changing distant "now".
I did not introduce any accelerated reference frames, which need additional definitions and the introduction of such is only good for creating more confusion when discussing SR. In the standard twin paradox the turn-around is of negligible duration so that the effect of acceleration on accumulated clock time of the accelerated clock is negligible, even if the so-called "clock hypothesis" were not applicable.

Einstein introduced the clock hypothesis as follows in his 1905 example:
"If we assume that the result proved for a polygonal line is also valid for a continuously curved line "
- end of section 4 of http://www.fourmilab.ch/etexts/einstein/specrel/www/

This was experimentally confirmed for muons at extreme accelerations.

 

[kamenjar]

You are only half correct. When the traveling twin departs at 90%c, they both do see each others clock equally slow--a factor of 0.2294 times their own. But this is true only for the outbound portion of the trip. Things are different for the inbound portion of the trip. As soon as the traveling twin turns around, he immediately sees the Earth twin's clock going fast--4.359 times his own. Since he spends an equal amount of time going out as coming in, you can easily calculate how much of a difference there will be in the amount the two twins aged by simply taking an average of the two factors. The average of 0.2294 and 4.359 is 2.2942 so however much the traveling twin aged during the trip, his Earth twin will age 2.2942 times as much. Simple, isn't it?

Then at what point does the clock discrepency occur?
1) During acceleration?
If so, then the difference of age is a function of acceleration. If so, then then it's wrong because we know that the longer the twin travels, the less he ages, and it has nothing to do with how long he accelerates.
2) During traveling?
If so, then your statement doesn't make sense to me if you are saying that they both see their clocks run equally fast or slow (it may make sense to someone else). The only way it would make sense is if they would both see their clocks run slower outbound and faster inbound but by different amounts.
Edir: Or are you saying that the Earth twin still sees the other twin's clock run slower during inbound trip?

By the way, I think that I understand relativistic principles, but I can't deal with formula answers.

 

[Jeronimus]

Things are different for the inbound portion of the trip. As soon as the traveling twin turns around, he immediately sees the Earth twin's clock going fast--4.359 times his own.

Where did you get the 4.359x from?

What the "turn around" aka acceleration towards the staying at home twin does is changing the frame the traveling twin is at rest.
Events that took place simultaneous in his former frame that are not local, will now be time shifted. The shift in time depends on the distance and the vrel to his former frame.

We usually assume instantaneous acceleration where the shift would happen instantly.

If the traveling twin was omniscient and he could watch the Earth's twin clock while doing a NON-instantaneous acceleration, then depending on the distance and acceleration he might very well see the Earth's twin clock go faster until the acceleration is done, but after that, while he is back on the way to earth, he will see the Earth twin's clocks go just as slow as the Earth twin sees his clock go slower.

You seem a bit confused about how this really works.

 

[ghwellsjr]

You are only half correct. When the traveling twin departs at 90%c, they both do see each others clock equally slow--a factor of 0.2294 times their own. But this is true only for the outbound portion of the trip. Things are different for the inbound portion of the trip. As soon as the traveling twin turns around, he immediately sees the Earth twin's clock going fast--4.359 times his own. Since he spends an equal amount of time going out as coming in, you can easily calculate how much of a difference there will be in the amount the two twins aged by simply taking an average of the two factors. The average of 0.2294 and 4.359 is 2.2942 so however much the traveling twin aged during the trip, his Earth twin will age 2.2942 times as much. Simple, isn't it?

well it surely is but isn't it said that a person sees the clock fastest which is at rest compared to him so can he really see the other clock running 4.359 (or whatever)
also is there droppler's effect also involved in this in any way?

Yes, the Doppler effect describes what each twin sees of the other ones clock. Isn't that what you asked about? (I'm assuming that the twins have special goggles or electronic equipment that permit them to see the red and blue shifted images.)

And yes, the traveling twin does see the Earth twin's clock running faster than his own on the return half of the trip, but you have to average that with him seeing it running slower during the first half of the trip and it's the average which tells us the final difference in aging.

I'm glad you agree that this is simple.

 

[Jeronimus]

Yes, the Doppler effect describes what each twin sees of the other ones clock. Isn't that what you asked about? (I'm assuming that the twins have special goggles or electronic equipment that permit them to see the red and blue shifted images.)

And yes, the traveling twin does see the Earth twin's clock running faster than his own on the return half of the trip, but you have to average that with him seeing it running slower during the first half of the trip and it's the average which tells us the final difference in aging.

I'm glad you agree that this is simple.

I see now. This is what happens if you take "see" literary instead of assuming the traveling twin actually knows physics and is able to calculate the real pace of the Earth twin's clock while traveling towards him, instead of him getting fooled by visual data creating the illusion of the Earth twin's clock is running faster because he is traveling towards the photons emitted by the Earth twin's clock.

 

[ghwellsjr]

Where did you get the 4.359x from?

It comes from the Relativistic Doppler Factor:

√((1+β)/(1-β)) = √((1+0.9)/(1-0.9)) = √((1.9)/(0.1)) = √19 ≈ 4.359

What the "turn around" aka acceleration towards the staying at home twin does is changing the frame the traveling twin is at rest.
Events that took place simultaneous in his former frame that are not local, will now be time shifted. The shift in time depends on the distance and the vrel to his former frame.

Have you actually done an analysis of the Twin Paradox by this method? Can you show us some numbers? Have you demonstrated that the traveling twin will see exactly what I described in post #5 or do you think your analysis will come up with something different? I want to see your numbers, please.

We usually assume instantaneous acceleration where the shift would happen instantly.

Yes, we do and so did I in post #5.

If the traveling twin was omniscient and he could watch the Earth's twin clock while doing a NON-instantaneous acceleration, then depending on the distance and acceleration he might very well see the Earth's twin clock go faster until the acceleration is done, but after that, while he is back on the way to earth, he will see the Earth twin's clocks go just as slow as the Earth twin sees his clock go slower.

But he's not omniscient and neither are we so we have to settle for what we can actually see and not some wishful thinking about remote viewing. If the acceleration is not instantaneous, it will only make the transistion of the traveling twin seeing the Earth twin's clock ticking at 0.2294 of his own to 4.359 of his own take a longer time instead of being instantaneous.

You seem a bit confused about how this really works.

Please work out the numbers and show the calculation for what the traveling twin actually sees and then maybe the confusion will subside.

 

[TrickyDicky]

There is a further difficulty the way I see it, it is assumed that the twin on Earth is inertial, while we know that due to its rotational and translational motions, is noninertial. But I guess that is part of the idealization of the problem.

 

[ghwellsjr]

I see now. This is what happens if you take "see" literary instead of assuming the traveling twin actually knows physics and is able to calculate the real pace of the Earth twin's clock while traveling towards him, instead of him getting fooled by visual data creating the illusion of the Earth twin's clock is running faster because he is traveling towards the photons emitted by the Earth twin's clock.

If the traveling twin actually knows physics, he would be aware that there is no such thing as the "real pace of the Earth twin's clock while traveling towards him". He would know that he can analyze the pace of both of their clocks from any inertial frame of reference and each one can assign different paces to their two clocks, none of which can be considered "real". What's real is the visual data that you call an illusion. Furthermore, each one of these inertial reference frames will agree on exactly what each twin sees throughout the entire trip. You can also analyze the scenario from non-inertial frames or jumping inertial frames and they can assign completely different paces to the two clocks but they will all agree on what each twin really sees.

 

[harrylin]

I see now. This is what happens if you take "see" literary instead of assuming the traveling twin actually knows physics and is able to calculate the real pace of the Earth twin's clock while traveling towards him, instead of him getting fooled by visual data creating the illusion of the Earth twin's clock is running faster because he is traveling towards the photons emitted by the Earth twin's clock.

Sorry, it's almost the contrary! What they literally see are true observations (measurements), which SR must be able to predict. And what they calculate based on arbitrary assumptions is in conflict with what others calculate based on different arbitrary assumptions; it's therefore erroneous (and leading to paradoxes!) to think that the traveling twin is able to calculate the real pace of the Earth twin's clock. If he knows SR, then he understands that he can not do such a thing. Different inertial reference systems assign different clock rates to that clock.

[ah I see that George was faster than me ]

 

[Dale]

Then at what point does the clock discrepency occur?

If you have a triangle then the sum of the lengths of two sides is greater than the length of the third side. At what point does this discrepancy occur?

 

[Jeronimus]

It comes from the Relativistic Doppler Factor:

√((1+β)/(1-β)) = √((1+0.9)/(1-0.9)) = √((1.9)/(0.1)) = √19 ≈ 4.359

Have you actually done an analysis of the Twin Paradox by this method? Can you show us some numbers? Have you demonstrated that the traveling twin will see exactly what I described in post #5 or do you think your analysis will come up with something different? I want to see your numbers, please.

This is the most misleading method to get someone to understand what really happens.

But he's not omniscient and neither are we so we have to settle for what we can actually see and not some wishful thinking about remote viewing.

Omniscient is the shortcut for placing an army of observers all along the frame the traveling twin is at rest in. The army of observers has synced clocks and writes down each event happening inside that frame including the time when Earth's clock is passing by them and the time it shows when passing by them (locally).
The twin takes the event data sent to him and composes an x/t diagram.
He then is able to check how long one second measured on the Earth twin's clock takes within his frame.
He will find that ANY clock which is within the Earth's frame at rest, will run at the SAME pace, slowed down by a factor of about 0,4359s.
Otherwise said. The clocks which are at rest in Earth's frame, advance ~0,4359s for every second passed within the traveling twin's frame.



According to you, the pace at which clocks which are at rest in Earth's frame seen from the traveling twin's perspective depends on if the clocks are in front or behind the twin. So someone who is behind the twin or in front, will come to different conclusions on the same clock. This is ridiculous.


As for the calculations:

I describe here,
https://www.physicsforums.com/showthread.php?t=640671

how to get Δt

And yes, i have done it. And yes, Δt is the same NO matter if you passed a clock or it is moving towards you...
Also, because of

"The laws of physics are the same in all inertial frames of reference."

clocks at rest in System A, traveling at vrel to System B will run slower by the same amount as clocks at rest in System B run slower in System A where they move at vrel.

 

[Jeronimus]

Sorry, it's almost the contrary! What they literally see are true observations (measurements), which SR must be able to predict. And what they calculate based on arbitrary assumptions is in conflict with what others calculate based on different arbitrary assumptions; it's therefore erroneous (and leading to paradoxes!) to think that the traveling twin is able to calculate the real pace of the Earth twin's clock. If he knows SR, then he understands that he can not do such a thing. Different inertial reference systems assign different clock rates to that clock.

[ah I see that George was faster than me ]

Not really. What george says is that the pace of a specific clock moving at vrel relative to an army of observers at rest within a given inertial reference system, depends on which of the observers you ask.

If that sounds good to you, so be it. But that is not really what is the case.(BTW - real was meant as WITHIN a given inertial frame of reference. Of course the pace is different for each different frame, but i never objected that. See above)

 

[ghwellsjr]

You are only half correct. When the traveling twin departs at 90%c, they both do see each others clock equally slow--a factor of 0.2294 times their own. But this is true only for the outbound portion of the trip. Things are different for the inbound portion of the trip. As soon as the traveling twin turns around, he immediately sees the Earth twin's clock going fast--4.359 times his own. Since he spends an equal amount of time going out as coming in, you can easily calculate how much of a difference there will be in the amount the two twins aged by simply taking an average of the two factors. The average of 0.2294 and 4.359 is 2.2942 so however much the traveling twin aged during the trip, his Earth twin will age 2.2942 times as much. Simple, isn't it?

Then at what point does the clock discrepency occur?
1) During acceleration?
If so, then the difference of age is a function of acceleration. If so, then then it's wrong because we know that the longer the twin travels, the less he ages, and it has nothing to do with how long he accelerates.
2) During traveling?
If so, then your statement doesn't make sense to me if you are saying that they both see their clocks run equally fast or slow (it may make sense to someone else). The only way it would make sense is if they would both see their clocks run slower outbound and faster inbound but by different amounts.
Edir: Or are you saying that the Earth twin still sees the other twin's clock run slower during inbound trip?

By the way, I think that I understand relativistic principles, but I can't deal with formula answers.

I didn't make any comment about what the Earth twin sees of the traveling twin's clock beyond the beginning of the trip but since you asked, I'll fill in the details.

The Earth twin will see the traveling twin's clock run slower than his own for much more than half the trip because he has to wait for the image of the distant turn around event to reach him over that long distance and this is (partly) what results in him seeing his twin's clock with less time on it when they finally get back together.

Another way of putting this is that he will see his twin traveling away from him for way more than half the time (while he sees his clock running slow) and then he sees him turn around and come back for a rather short time (while he sees his clock running fast). So I wouldn't say that the Earth twin sees the other twin's clock run slower during the inbound trip, it's that he sees the outbound trip last longer than the inbound trip. Remember, when he sees outbound, he sees a slow clock, when he sees inbound, he sees a fast clock.

How do you like that? No formulas!

 

[Jeronimus]

Just to make sure, i will explain how I would measure the pace of a moving clock within a system i am at rest in...Two observers which are at a distance x. They are both at rest to each other. They both synced their clocks.

A moving clock passes by the front observer, which writes down the time it passed by him AND the time the moving clock shows. To keep it simple, assume it was at t1=0s for the observer's clock and clockt1=0s.

The clock keeps moving at vrel and reaches the second observer at t2=10s. At vrel = 0.5c and a distance of 5 lightseconds, the moving clock at the second observer would display clockt2 ~8.66s.

Are we in agreement with this?
Are we in agreement that any two observers at a distance of 5 lightseconds within the SAME inertial frame of reference would observe the same if a similar clock moved by them at vrel = 0.5c?The two observers would sent each other the data, allowing them to conclude the pace of the moving clock.

 

[kamenjar]

I think that this whole thread and a multitude of others are here on this forum because some of the GR/SR are simply "stubborn" when making claims. Stubborn in a sense that some claims keep repeating all over posts and texts that are not true when all facts are accounted for.

To my understanding (and I may be wrong), there is no twin paradox and there's nothing paradoxical about it. The only paradoxical thing is the lame statement that says something about "both travelers agreeing that other person's CLOCK runs slower". That is just plain absurd. You can not make statement about the other person's clock when you are moving by just observing the rate of the ticks measured by your own clock rate.

The only way to make a statement about the other twin's CLOCK is to account for redshift. Then you can actually make statements about which clock runs slower and I believe that if someone (more capable than myself) that did calculations conclude that the traveling twin's clock is slower during the whole trip AND that both the traveler and the guy on Earth can do those measurements and calculations and conclud the same - the traveling twin's clock runs slower. So as ghwellsjr said, it's probably simple and there's nothing paradoxical or confusing.

 

[ghwellsjr]

This is the most misleading method to get someone to understand what really happens.

Ok, we'll agree to use your definition of what real means:

(BTW - real was meant as WITHIN a given inertial frame of reference. Of course the pace is different for each different frame, but i never objected that. See above)

Omniscient is the shortcut for placing an army of observers all along the frame the traveling twin is at rest in.

You just said that real meant within a given inertial frame of reference and now you want to talk about a frame that the traveling twin is a rest in. But it cannot be an inertial frame for the entire trip so how does that work?

The army of observers has synced clocks and writes down each event happening inside that frame including the time when Earth's clock is passing by them and the time it shows when passing by them (locally).
The twin takes the event data sent to him and composes an x/t diagram.
He then is able to check how long one second measured on the Earth twin's clock takes within his frame.
He will find that ANY clock which is within the Earth's frame at rest, will run at the SAME pace, slowed down by a factor of about 0,4359s.
Otherwise said. The clocks which are at rest in Earth's frame, advance ~0,4359s for every second passed within the traveling twin's frame.

I don't know why you want to make this so complicated. Let's do what you said and pick as our given inertial frame of reference the one in which the Earth twin is at rest and in which the traveling twin starts out and ends up at rest. In this frame the Earth twin's clock runs normally.

Now the traveling twin accelerates instantly to a speed of 90%c. Gamma at this speed is 2.294 (not 0.4359 as you claim in your linked diagram). That means that a clock traveling at 90%c will run slower by a factor of 2.294. The traveling twin's clock will run slower than the Earth twin's clock by this amount during his entire trip so when he gets back the Earth clock has advanced by 2.294 times whatever his clock advanced. This is what really happens according to your definition of real and it's exactly what I said would happen in post #5 and so I don't know why you say it's misleading.

According to you, the pace at which clocks which are at rest in Earth's frame seen from the traveling twin's perspective depends on if the clocks are in front or behind the twin. So someone who is behind the twin or in front, will come to different conclusions on the same clock. This is ridiculous.

It would be ridiculous if I ever said that. Can you point to the post where I said that?

As for the calculations:

I describe here,
https://www.physicsforums.com/showthread.php?t=640671

how to get Δt

And yes, i have done it. And yes, Δt is the same NO matter if you passed a clock or it is moving towards you...
Also, because of

"The laws of physics are the same in all inertial frames of reference."

clocks at rest in System A, traveling at vrel to System B will run slower by the same amount as clocks at rest in System B run slower in System A where they move at vrel.

But what I asked you to do was calculate what the traveling twin sees of the Earth's twin's clock and you haven't done that.

 

[marty1]

I definitely do not agree that velocity has any anything to do with this effect. Anyone in the universe can claim they are at rest so long as they are not accelerating. Even with constant acceleration they could claim they are in a gravitational field.

We can measure the difference in the speed of clocks on the surface of the Earth compared to those in orbit solely due to the affects of gravity. This is not disputed.

 

[ghwellsjr]

I definitely do not agree that velocity has any anything to do with this effect. Anyone in the universe can claim they are at rest so long as they are not accelerating. Even with constant acceleration they could claim they are in a gravitational field.

But the twins cannot both claim they continue at rest if they separate and then end up together at rest again.

What are you talking about?

We can measure the difference in the speed of clocks on the surface of the Earth compared to those in orbit solely due to the affects of gravity. This is not disputed.

We can measure the difference in the speed of clocks at different elevations on the surface of the Earth due just to the effects of gravity but if you're going to put one in orbit you have to also take into account its velocity. Your comment is disputed.

 

[ghwellsjr]

Just to make sure, i will explain how I would measure the pace of a moving clock within a system i am at rest in...Two observers which are at a distance x. They are both at rest to each other. They both synced their clocks.

A moving clock passes by the front observer, which writes down the time it passed by him AND the time the moving clock shows. To keep it simple, assume it was at t1=0s for the observer's clock and clockt1=0s.

The clock keeps moving at vrel and reaches the second observer at t2=10s. At vrel = 0.5c and a distance of 5 lightseconds, the moving clock at the second observer would display clockt2 ~8.66s.

Are we in agreement with this?
Are we in agreement that any two observers at a distance of 5 lightseconds within the SAME inertial frame of reference would observe the same if a similar clock moved by them at vrel = 0.5c?The two observers would sent each other the data, allowing them to conclude the pace of the moving clock.

What you are saying is correct.

However, what I want you to consider is that the two observers do not have to overtly send any data to the other observers, they can just look at it. EDIT: I went back and see that you want them to send both their own time and the time they see on the other clock, so yes, they do have to overtly send the time on the other clock.

So what I'm asking you to do is figure out how each of the clocks observes the time on each of the other clocks in your scenario.

 

[marty1]

But the twins cannot both claim they continue at rest if they separate and then end up together at rest again.

What are you talking about?

While of course they can resolve that after the fact.

We can measure the difference in the speed of clocks at different elevations on the surface of the Earth due just to the effects of gravity but if you're going to put one in orbit you have to also take into account its velocity. Your comment is disputed.

But the relative difference in speeds of the 2 clocks is zero for a geosynchronous clocks. They are always the same distance from one another. How fast is a clock in geosynchronous orbit moving away from you? It is not moving away from you. My satellite dish is always pointing in the same direction. The difference is the acceleration. Each of the clocks are accelerating toward the Earth at different rates and in effect accelerating around a different curve.

Unless, of course, you want to say that "absolute" speed matters.

I am just wondering now how fast a clock ticks if it follows the Earth in orbit around the sun.

Another point:

When the twins are moving away from one another they are advancing ahead of the beam of light that is the other person observation of the other clock so it will advance slower because you are always getting a little farther away (I think I said that right). If you could move at the speed of light the other clock would appear to stop. If you could move faster the other clock would move backwards.

 

[ghwellsjr]

While of course they can resolve that after the fact.

Resolve what? I don't know what you're talking about.

But the relative difference in speeds of the 2 clocks is zero for a geosynchronous clocks. They are always the same distance from one another. How fast is a clock in geosynchronous orbit moving away from you? It is not moving away from you. My satellite dish is always pointing in the same direction. The difference is the acceleration. Each of the clocks are accelerating toward the Earth at different rates and in effect accelerating around a different curve.

Unless, of course, you want to say that "absolute" speed matters.

In this sense, what you are calling "absolute" speed does matter. The satellites are traveling much faster than you are even though they appear to be stationary in the sky above you and you can't ignore their speed when calculating how much slower they are compared to you

I am just wondering now how fast a clock ticks if it follows the Earth in orbit around the sun.

I don't really know but again, you can't ignore its speed.

Another point:

When the twins are moving away from one another they are advancing ahead of the beam of light that is the other person observation of the other clock so it will advance slower because you are always getting a little farther away (I think I said that right). If you could move at the speed of light the other clock would appear to stop. If you could move faster the other clock would move backwards.

You can't move at the speed of light but you can get as close as you want and the clock will be running as slow as you want (but not stopped). And you can't go faster so it's pointless to say what it would do.

 

[Dale]

To my understanding (and I may be wrong), there is no twin paradox and there's nothing paradoxical about it.

I agree. It isn't a true paradox. It is more like a very famous homework problem that introductory students often get wrong.

The only paradoxical thing is the lame statement that says something about "both travelers agreeing that other person's CLOCK runs slower". That is just plain absurd.

The statement isn't absurd, but it is a little off. It should read, "both travelers agree that the other person's clock runs slower in their reference frame". How fast a clock runs is a frame dependent quantity, so the frame needs to be specified.

I believe that if someone (more capable than myself) that did calculations conclude that the traveling twin's clock is slower during the whole trip

Again, how fast a clock runs is frame dependent. Your belief is true in some frames, but false in others.

 

[marty1]

Resolve what? I don't know what you're talking about.

In this sense, what you are calling "absolute" speed does matter. The satellites are traveling much faster than you are even though they appear to be stationary in the sky above you and you can't ignore their speed when calculating how much slower they are compared to you

I don't really know but again, you can't ignore its speed.

You can't move at the speed of light but you can get as close as you want and the clock will be running as slow as you want (but not stopped). And you can't go faster so it's pointless to say what it would do.

My point is that seeing the clocks moving slower is NOT due to time dilation but the fact that you are not seeing the light pass you from a fixed point but always seeing it later and later with the extreme of going toward stopping and even reversing... all the while there was no time warping or dilation involved.

 

[zonde]

Acceleration have coordinate effect on clock reading. That's because acceleration changes the hyperplane of now.
Imagine two observers at the same spot moving at the same speed in respect to Earth but in opposite directions (toward/away). While they see exactly the same picture of Earth they will assign different readings to the "now" of Earth clock.

Only if and when they switch the inertial reference frames that they use, as I mentioned. For example the astronauts in the international space-station likely stick to the ECI frame instead of continuously changing distant "now".
I did not introduce any accelerated reference frames, which need additional definitions and the introduction of such is only good for creating more confusion when discussing SR. In the standard twin paradox the turn-around is of negligible duration so that the effect of acceleration on accumulated clock time of the accelerated clock is negligible, even if the so-called "clock hypothesis" were not applicable.

Probably I was not clear enough about what I was talking.
The effect applies to remote (Earth) clock, not local clock. So this has nothing to do with local clock and accumulated clock time.
When rocket has finished it's turn-around then Earth clock has made a "jump" ahead. This "jump" ahead is a coordinate effect.

And my example was intended to show that we can avoid acceleration part but the effect is still there. Imagine that the twin that is heading away from Earth instead of turning back pass another astronaut that is heading toward Earth. And when they pass each other they simply exchange their clock readings. But they will disagree what time it is on Earth now and that is the same effect as the one of changing frames (accelerating).